- Platinum cannot be plated from solution by electrolysis. True
- Reactions which are readily reversed by a small change in concentration are ones in which E is close to zero. True
- All solutions of electrolytes conduct electricity. True ( varying degrees), non electrolytes do not
- Electrolysis can only be performed on solutions. False
- In electrolysis, the process occurring at the cathode is oxidation. False
- In any cell, electrolytic or voltaic, the cathode is the negative electrode. False
- An anode is a always the positive electrode. False
- The liberation of free elements from liquids by the passage of an electric current is called neutralization. (F) Called discharging
- Electrolysis can be carried out in molten salts as well as in water solutions. True
- Electroplating is the reverse of electrolysis. False
9. Consider the cell: Ag(s) | Ag+(1.0M) || Cu2+(1.0M) | Cu(s)
a.) Write the chemical reaction that takes place in this cell. In which direction will the reaction proceed spontaneously?
Ag+ + e- → Ag(s) E°red = +0.800 V
Cu2+ + e- → Cu(s) E°red = +0.340 V
Remember that the reaction with the larger or more positive reduction potential will be the one more likely to be reduced. Looking at the above cell diagram (always written with anode first, then cathode), the following reaction would occur:
2Ag(s) + Cu2+ → 2Ag+ + Cu(s)
- Anode Reaction: Oxidation (2Ag(s) → 2Ag+ + 2e-)
- Cathode Reaction: Reduction (Cu2+ + 2e- → 2Cu(s))
But this is not the spontaneous reaction since we know that the Ag+/Ag is the species most likely to be reduced. The spontaneous reaction must be the reverse reaction:
2Ag+ + Cu(s) → 2Ag(s) + Cu2+
- Anode Reaction: Oxidation (Cu(s) → Cu2+ + e-)
- Cathode Reaction: Reduction (2Ag+ + 2e- → 2Ag(s))
E°cell = E°cathode - E°anode = E°red, Ag - E°red, Cu = 0.800V – 0.340 V = +0.460 V
c.) Do electrons flow from Ag to Cu in the external circuit, or the other way?
The electrons always flow from the anode to the cathode. Since we discovered that the Cu is the anode and the Ag is the cathode, the electrons will flow the other way from the Cu to the Ag.
10. Consider the following cell: Ni | Ni2+(0.010M) || Sn2+(1.0M) | Sn
- Predict the direction in which spontaneous reactions will occur.
- Which metal, Ni or Sn, will be the cathode and which the anode?
- What is Eo for the cell?
- What will E be for the cell with the specified concentrations at 25°C?
11. Consider the cell: Sn | SnCl2 (0.10M) || AgCl(s) || Ag
- Will electrons flow spontaneously from Sn to Ag, or in the reverse direction?
- What is the standard potential, Eo, for the cell?
- What will the cell potential, E, be at 25°C?
Part a.) Predict the way of the electrons flow. Is Eocell positive or negative? Positive Eocell indicates spontaneous reaction.
Where: Eocell = Eocathode - Eoanode = Eoright - Eoleft
Eoanode = Sn0 → Sn2+ +e- from SRP reference table = -0.137 V
Eocathode= AgCl(s) + e- → Ag(s) + Cl- from SRP reference table = 0.2223 V
Note: The silver/silver chloride or Ag/AgCl reference electrode is a common reference electrode. It is easily and cheaply prepared and is stable, and very robust. It is sometimes referred to as "SSCE" (Silver/Silver Chloride Electrode).
Eocell = 0.2223 V - -0.137 V = 0.3593 V
Part b) Eocell = 0.2223 V - -0.137 V= 0.3593 V
Part c) Using Nernst equation 20.18 (pg. 837)
Ecell = Eocell - (0.0592/n)log Q
Need to know chemical reaction…. Electrons move from Sn to Ag so the chemical equation is:
Sn0 + 2Ag+ → Sn2+ + Ag0
n is the number of electrons involved per reaction, in this reaction n=2
Q the reaction quotient is equal to [products]/ [reactants]
The initial conditions the reactants are solids and thus are excluded from this equal because it is less mobile. So the quotient is equal to only the products which is 0.10M.
Ecell = 0.3593V - (0.0592/2) log 0.10 = 0.39 V
12. Use the line notation of the previous problems to represent a cell that uses the following half-reactions:
PbO2 + 4H+ + 2e- → Pb2+ + 2H2O
PbSO4 + 2e- → Pb + SO42-
- Which is the reaction at the cathode of the cell? Which way do electrons flow in an external circuit?
- What is Eo for this cell?
Pb(s), PbSO4(s) || PbO2(s) | Pb2+(aq)
a) reduction reaction occurs at the cathode (PbO2 + 4H+ + 2e- → Pb2+ + 2H2O). electrons will flow from the cathode to the anode.
b) Ecell = Ecathode - Eanode = 1.468 - (-0.3588) = 1.8268 V
14. Find the missing standard reduction potentials for these following half-reactions:
|MnO4- + 8H+ + 5e- → Mn2+ + 4H2O||+ 1.49|
|Au3+ + 3e- → Au(s)||+ 1.42|
|Cl2 + 2e- → 2Cl-||1.36 V|
|AuCl4- + 3e- → Au(s) + 4Cl-||1.002 V|
|4H+ + NO3- + 3e- → NO + 2H2O||0.96 V|
lf we assume that all reactants and products are at unit activity:
- Which substance in the half-reactions given is the best oxidizing agent? Which is the best reducing agent?
- Will permanganate oxidize metallic gold?
- Will metallic gold reduce nitric acid?
- Will nitric acid oxidize metallic gold in the presence of Cl- ion?
- Will metallic gold reduce pure Cl2 gas in the presence of water?
- Will chlorine oxidize metallic gold if Cl- ion is present?
- Will permanganate oxidize chloride ions?
Cl2/Cl- Eo = 1.36 V
AuCl4-/Au Eo = 0.93 V
NO3-/NO Eo = 0.96 V
a. MnO4- is the best oxidizing agent; Au(s) (in a Cl- environment) is the best reducing agent.
Hint: Aqua regia (concentrated nitric acid and concentrated hydrochloric acid) dissolves gold, though neither constituent acid will do so alone, because, in combination, each acid performs a different task. Nitric acid is a powerful oxidizer, which will actually dissolve a virtually undetectable amount of gold, forming gold ions (Au3+). The hydrochloric acid provides a ready supply of chloride ions (Cl-), which react with the gold ions to produce chloroaurate anions, also in solution. The reaction with hydrochloric acid is an equilibrium reaction which favors formation of chloroaurate anions (AuCl4-). This results in a removal of gold ions from solution and allows further oxidation of gold to take place. The gold dissolves to become chloroauric acid. In addition, gold may be dissolved by the free chlorine present in aqua regia. Appropriate equations are
Au (s) + 3 NO3- (aq) + 6 H+ (aq) → Au3+ (aq) + 3 NO2 (g) + 3 H2O (l) and
Au3+ (aq) + 4 Cl- (aq) → AuCl4- (aq).
The oxidation reaction can also be written with nitric oxide as the product rather than nitrogen dioxide:
Au (s) + NO3- (aq) + 4 H+ (aq) → Au3+ (aq) + NO (g) + 2 H2O (l).
16. Find the standard reduction potentials for the following half-reactions;
SO42- + 4H+ + 2e- → H2SO3 + H2O
Ag+ + e- → Ag
- Write the balanced overall reaction for a successful cell made from these two couples,
- Write the line notation for the cell.
- What is E° for the cell?
- What is the equilibrium constant for the cell reaction at 25°C?
- Calculate the ratio of activities of products and reactants, Q, that will produce a cell voltage of 0.51 V.
1. The balanced half reactions and their Reduction Potentials (from SRP reference table) will be:
SO42- + 4H+ + 2e- → H2SO3 + H2O E = +0.1576
2Ag+ + 2e- → 2Ag(s) E = +.799
Because the Ag(s)/Ag+ reaction is more positive (stronger) it will run in the forward (reduction) direction. To balance, H2SO3/SO42- will run in the opposite (Oxidation) direction. The net equation becomes:
H2SO3 + H2O + 2Ag+ → SO42- + 4H+ + 2Ag(s)
2. H2SO3 | SO42- || Ag+ | Ag(s)
3. E° = E°reduction + E°oxidation = +0.799 + (-0.1576) = +0.6414 V
(where E°oxidation is the reverse of the reduction potential for the oxidation reaction)
Or written another way
E° = E°right - E°Left which is the same as E° = E°cathode - E°anode
E° = +0.799 - (+0.1576) = +0.6414 V
(where E°cathode the reduction potential of the reaction at the cathode, and E°anode = the reduction potential of the reaction at the anode. No values have been reversed)
4. We can solve this question using the Nernst equation. E = E° - (RT/nF)lnQ.
At 25°C the Nernst equation can be written in terms of log as E = E° - (0.059/n)logQ
At equilibrium E = 0, and Q = Keq which gives: 0 = E° - (0.059/n)logKeq
Keq = 10^(nE°/0.059)
For the system described in this problem n = (# of electrons) = 2
Keq = 10^[(2*0.6414)/0.059)] = 5.5 x 1021
5. Using the equation E = E° - (0.059/n)logQ, with E = 0.51 V gives:
0.51 = 0.6414 - (0.059/2)logQ
Q = 10^[((0.51-0.6414)2)/-0.059] = 2.8 x 104
In this equation, only the REDUCTION potentials are used for both the oxidation and reduction half reaction. E°cell > 0, which is what we would expect for a spontaneous reaction.